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1. Area and Co-Area Formula 1.1. Hausdorff Measure. in This Section 1. Area and co-area formula 1.1. Hausdorff measure. In this section we will recall the definition of the Hausdorff measure and we will state some of its basic properties. A more detailed discussion is postponed to Section ??. s=2 s Let !s = π =Γ(1 + 2 ), s ≥ 0. If s = n is a positive integer, then !n is volume of the 1 unit ball in Rn. Let X be a metric space. For " > 0 and E ⊂ X we define 1 !s X Hs(E) = inf (diam A )s " 2s i i=1 where the infimum is taken over all possible coverings 1 [ E ⊂ Ai with diam Ai ≤ ": i=1 s Since the function " 7! H"(E) is nonincreasing, the limit s s H (E) = lim H"(E) "!0 exists. Hs is called the Hausdorff measure. It is easy to see that if s = 0, H0 is the counting measure. s P1 s The Hausdorff content H1(E) is defined as the infimum of i=1 ri over all coverings 1 [ E ⊂ B(xi; ri) i=1 s of E by balls of radii ri. It is an easy exercise to show that H (E) = 0 if and only if s H1(E) = 0. Often it is easier to use the Hausdorff content to show that the Hausdorff measure of a set is zero, because one does not have to worry about the diameters of the sets in the covering. The Hausdorff content is an outer measure, but very few sets are s actually measurable, and it is not countably additive on Borel sets. This is why H1 is called content, but not measure. Theorem 1.1. Hs is a metric outer measure i.e. Hs(E [ F ) = Hs(E) + Hs(F ) whenever E and F are arbitrary sets with dist (E; F ) > 0. Hence all Borel sets are Hs measurable. It is an easy exercise to prove that Hs is an outer measure. The fact that it is a metric outer measure follows from the observation that if " < dist (E; F )=2, we can assume that 1 n !n n s If B ⊂ R is a ball, then 2n (diam B) = jBj. This explains the choice of the coefficient !s=2 in the definition of the Hausdorff measure. 1 2 sets of diameter less than " that cover E are disjoint from the sets of diameter less than " that cover F . We leave details as an exercise. Finally measurability of Borel sets is a general property of metric outer measures. The next result is very important and difficult. We will prove it in Section ?? Theorem 1.2. Hn on Rn coincides with the outer Lebesgue measure Ln. Hence a set is Hn measurable if and only if it is Lebesgue measurable and both measures are equal on the class of measurable sets. This result generalizes to the case of the Lebesgue measure on submanifolds of Rn. We will discuss it in the Subsection 1.3. In what follows we will often use the Hausdorff measure notation to denote the Lebesgue measure. Proposition 1.3. If f : X ⊃ E ! Y is a Lipschitz mapping between metric spaces, then Hs(f(E)) ≤ LsHs(E). In particular if Hs(E) = 0, then Hs(f(E)) = 0. This is very easy. Indeed if A ⊂ E, then f(A) has diameter less than or equal to Ldiam A, where L is the Lipschitz constant of f. This observation and the definition of the Hausdorff measure easily yields the result. In particular, if f : Rn ⊃ E ! Rm is a Lipschitz mapping and jEj = 0, then Hn(f(E)) = 0. We will prove a stronger result which is known as the Sard theorem. A more general version of the Sard theorem will be discussed in Section ??. Theorem 1.4 (Sard). Let f : Rn ⊃ E ! Rm be Lipschitz continuous and let Crit (f) = fx 2 E : rank apDf(x) < ng; then Hn(f(Crit (f))) = 0. In the proof we will need the so called 5r-covering lemma. It is also called a Vitali type covering lemma. Here and in what follows by σB we denote a ball concentric with the ball B and σ times the radius. Theorem 1.5 (5r-covering lemma). Let B be a family of balls in a metric space such that supfdiam B : B 2 Bg < 1. Then there is a subfamily of pairwise disjoint balls B0 ⊂ B 3 such that [ [ B ⊂ 5B: B2B B2B0 If the metric space is separable, then the family B0 is countable and we can arrange it as a 0 1 sequence B = fBigi=1, so 1 [ [ B ⊂ 5Bi : B2B i=1 Remark 1.6. Here B can be either a family of open balls or closed balls. In both cases the proof is the same. Proof. Let supfdiam B : B 2 Bg = R < 1. Divide the family B according to the diameter of the balls R R F = fB 2 B : < diam B ≤ g : j 2j 2j−1 S1 Clearly B = j=1 Fj. Define B1 ⊂ F1 to be the maximal family of pairwise disjoint balls. Suppose the families B1;:::; Bj−1 are already defined. Then we define Bj to be the maximal family of pairwise disjoint balls in j−1 0 0 [ Fj \ fB : B \ B = ; for all B 2 Big : i=1 0 S1 Next we define B = j=1 Bj. Observe that every ball B 2 Fj intersects with a ball in Sj Sj i=1 Bj. Suppose that B \ B1 6= ;, B1 2 i=1 Bi. Then R R diam B ≤ = 2 · ≤ 2 diam B 2j−1 2j 1 and hence B ⊂ 5B1. The proof is complete. Proof of the Sard theorem. Using the McShane extension (Theorem ??) we can assume that f is defined on all of Rn and replace the approximate derivative by the classical one. Indeed, the set of points in E where the approximate derivative exists, but the extension to Rn is not differentiable at these points has measure zero and this set is mapped onto a set of Hn measure zero. Let Z be the set of points in Rn such that Df(x) exists and rank Df(x) < n. We need to show that Hn(f(Z)) = 0. By splitting Z into bounded pieces we may assume that Z is 2 contained in the interior of the unit cube Q. For L > " > 0 and x 2 Z there is rx > 0 2Indeed, if each bounded piece of Z is mapped into a set of Hn measure zero, then Z is mapped into a set of measure zero. 4 such that B(x; rx) ⊂ Q and jf(y) − f(x) − Df(x)(y − x)j < "rx if y 2 B(x; 5rx). Hence dist (f(y);Wx) ≤ "rx for y 2 B(x; 5rx), n where Wx = f(x) + Df(x)(R ) is an affine space through f(x). Clearly dim Wx ≤ n − 1. Thus (1.1) f(B(x; 5rx)) ⊂ B(f(x); 5Lrx) \ fz : dist (z; Wx) ≤ "rxg: Since dim Wx = k ≤ n − 1 we have that n n−1 n H1(f(B(x; 5rx)) ≤ C"L rx ; where the constant C depends on n only. Indeed, the k dimensional ball B(f(x); 5Lrx)\Wx can be covered by Lr k Ln−1 C x ≤ C "rx " 3 balls of radius "rx. Then balls with radii 2"rx and the same centers cover the right hand side of (1.1). Thus Ln−1 Hn (f(B(x; 5r )) ≤ C (4"r )n = C0"rnLn−1: 1 x " x x S From the covering Z ⊂ x2Z B(x; rx) we can select a family of pairwise disjoint balls S B(xi; rxi ), i = 1; 2;::: such that Z ⊂ i B(xi; 5rxi ). We have 1 1 X X Hn (f(Z)) ≤ Hn (f(B(x ; 5r )) ≤ C"Ln−1 rn ≤ C0"Ln−1; 1 1 i xi xi i=1 i=1 because the balls B(xi; rxi ) are disjoint and contained in the unit cube; hence the sum of their volumes is less than one. Since " can be arbitrarily small we conclude that n n H1(f(Z)) = 0 and thus H (f(Z)) = 0. Exercise 1.7. Show that if •Hs(E) < 1, then Ht(E) = 0 for all t > s ≥ 0; •Hs(E) > 0, then Ht(E) = 1 for all 0 ≤ t < s. 3 and hence diameter 4"rx 5 Definition 1.8. The Hausdorff dimension is defined as follows. If Hs(E) > 0 for all s ≥ 0, then dimH (E) = 1. Otherwise we define s dimH (E) = inffs ≥ 0 : H (E) = 0g: It follows from the exercise that there is s 2 [0; 1] such that Ht(E) = 0 for t > s and Ht(E) = 1 for 0 < t < s. Hausdorff dimension of E equals s. It also easily follows from Proposition 1.3 that Lipschitz mappings do not increase the Hausdorff dimension. 1.2. Countably rectifiable sets. Definition 1.9. We say that a metric space X is countably n-rectifiable if there is a family n of Lipschitz mappings fi : R ⊃ Ei ! X defined on measurable sets such that 1 ! n [ H X n f(Ei) = 0: i=1 In particular we can talk about sets X ⊂ Rm that are countably n-rectifiable. Clearly any Borel subset of a countably n-rectifiable set is countably n-rectifiable. In other words X is countably n-rectifiable if it can be covered by countably many Lipschitz images of subsets of Rn up to a set of Hn measure zero. Since Lipschitz mappings map sets of finite Hn measure onto sets of finite Hn measure, the Hn measure on X is σ-finite and hence dimH X ≤ n.
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